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Does anyone know why $$\frac C{1+r}+\frac C{(1+r)^2}+\frac C{(1+r)^3}+\cdots=\frac Cr?$$

It probably has to do with power basics, but how to solve it.

user246336
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user274094
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4 Answers4

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Hint

Geometric progression gives you :

$1+ \frac{1}{x} + \frac{1}{x^2} + ... = \frac{1}{1-x}$

Take $c$ in front of the expression and you're left with: $$c \bigg(\frac{1}{1+r} + \frac{1}{(1+r)^2} + ... \bigg)$$

I think you can see the solution from here.

Transcendental
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$$\require{color}\frac{C}{1+r}+\frac{C}{(1+r)^2}+\frac{C}{(1+r)^3}+\cdots = {C\color{red}\left(\frac{1}{1+r}+\frac{1}{(1+r)^2}+\frac{1}{(1+r)^3}+\cdots\right)}=\boxed{C\color{red}{x}}$$

Then,

$$(r+1)C{\color{red}x}=C\left(1+\underbrace{{\color{red}\frac1{r+1}+\frac1{(r+1)^2}+\cdots}}_{x}\right) $$

This is,

$$(r+1){\color{red}x} = 1+{\color{red}x}\implies {\color{red}x} = \frac1r$$

Hence,

$$\boxed{C{\color{red}x}} = \frac Cr$$

EDIT: solving equation $(r+1){\color{red}x} = 1+{\color{red}x}$.

\begin{align} \require{cancel} (r+1){\color{red}x}& = 1+{\color{red}x}\\ r\cdot{\color{red}x}+1\cdot{\color{red}x}& = 1+{\color{red}x}\\ r\cdot{\color{red}x}+{\color{red}x}& = 1+{\color{red}x}\\ r\cdot{\color{red}x}+\cancel{{\color{red}x}}& = 1+\cancel{{\color{red}x}}\\ r\cdot{\color{red}x}& = 1\\ {\color{red}x}& = \frac1r\\ \end{align}

user246336
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  • Thanks, what is Cx? what does it represent? BTW Is this a financial source that simplify this kind of formulas? – user274094 Sep 25 '15 at 18:46
  • $C$ multiplied by $x$. $x$ is defined at the sum in the answer. This is a standard trick for summing an infinite geometric series – Dr Xorile Sep 25 '15 at 18:54
  • Also this answer is only valid for $r>0$ or $r<-2$. – Dr Xorile Sep 25 '15 at 18:55
  • OK, what happened in the 3rd and fourth step? – user274094 Sep 25 '15 at 19:04
  • The third step is the same as the second one but with the $x$ substituted. Then, solving for $x$ we get the value of the infinite sum. In the forth step, your original infinite sum is equal to $Cx$, by the definition of $x$, and thus equal to $C$ times the value of $x$, this is, $1/r$. – user246336 Sep 25 '15 at 19:26
  • I've edited for more clarity. Which is in red is always $x$. And what is in the boxes is what you aim to know. – user246336 Sep 25 '15 at 19:32
  • It's all clear except for one step, solving 1+x = x(1+r) will lead to x=1+x/1+r, can you explain how would that equal 1/r? – user274094 Sep 25 '15 at 19:41
  • I've included a step-by-step solution of the equation. – user246336 Sep 25 '15 at 19:47
  • Thanks iqcd, it works now. It's a way better to understand how to solve things instead of memorizing. Cheers. – user274094 Sep 25 '15 at 19:51
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Hint:

Multiply the series by $1+r$ and subtract $C$. What do you get ?

  • This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. – daOnlyBG Sep 25 '15 at 21:19
  • @daOnlyBG: this is clearly flagged as a hint. This is very common practice on Mathematics, isn't it ? The hint by Transcendental even says less. –  Sep 26 '15 at 08:08
  • yes, you are correct. Originally my impression was that short hints were supposed to be left in the comments, but this thread clarified it quite well: http://meta.math.stackexchange.com/q/10589/173397 I'll up vote your post – daOnlyBG Sep 26 '15 at 17:02
  • @daOnlyBG: thanks. I use to produce short/simple answers (hopefully not too short), as I believe there is value in brevity. I am "penalized" every now and then for this. –  Sep 27 '15 at 17:23
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Let $ 1+r = k $ so we want to show that $$\sum_{i=1}^{i=\infty}\frac{c}{k^{i}}=\frac{c}{r}$$ We sum of a geometric series which can be solved as, $$\frac{G}{1-R}$$, where G is the first term and R is the ratio. So we get $$ \frac{c}{k(1-\frac{1}{k})}$$ $$ \frac{c}{(k-1)}$$ $$\frac{c}{r}$$

Socre
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